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Question: Two-point charges \[{{q}_{A}}=5\mu C\]and \[{{q}_{B}}=-5\mu C\] are located at A and B separated by ...

Two-point charges qA=5μC{{q}_{A}}=5\mu Cand qB=−5μC{{q}_{B}}=-5\mu C are located at A and B separated by 0.2 m in vacuum.
A. What is the electric field at the midpoint O of the line joining the two charges?
B. If a negative test charge of magnitude 2 nC is placed at the midpoint, what is the force experienced by it?

Explanation

Solution

We are given two charges which are equal in magnitude but opposite in polarity and separated by 0.2m distance. We know two charges which are equal in magnitude but opposite in polarity separated by some distance forms an electric dipole. We need to keep in mind that force is a vector quantity. It has both the magnitude and the direction.

Complete step by step answer:
A. The electric field is given by the formula: E=kqAr2E=k\dfrac{{{q}_{A}}}{{{r}^{2}}}.
Now for both the charges, r= 0.1m
For first charge, E→A=kqAr2n^{{\overrightarrow{E}}_{A}}=k\dfrac{{{q}_{A}}}{{{r}^{2}}}\widehat{n}, the charge is positive so field is directed outwards.
For second charge, E→B=kqBr2n^{{\overrightarrow{E}}_{B}}=k\dfrac{{{q}_{B}}}{{{r}^{2}}}\widehat{n}, the charge is negative, so field is directed inwards
Now the magnitude of charges is same that is qA=qB=q{{q}_{A}}={{q}_{B}}=q
The net field thus both in same direction, gets added up and so the net field is E→=2kqr2n^\overrightarrow{E}=2k\dfrac{q}{{{r}^{2}}}\widehat{n}
⇒E→=2×9×109×5×10−60.12n^ ∴E→=9×106n^N/C\Rightarrow \overrightarrow{E}=2\times 9\times {{10}^{9}}\times \dfrac{5\times {{10}^{-6}}}{{{0.1}^{2}}}\widehat{n} \\\ \therefore \overrightarrow{E}=9\times {{10}^{6}}\widehat{n}N/C

Hence,the electric field at the midpoint O of the line joining the two charges is 9×106n^N/C9\times {{10}^{6}}\widehat{n}N/C.

B. Now a charge of 2 nC is placed at the midpoint and we know the value of electric field at midpoint is 9×106n^N/C9\times {{10}^{6}}\widehat{n}N/C, so force is given by the formula, F=qE
⇒F=qE ⇒F=2×10−9×9×106 ∴F→=0.018n^N/C\Rightarrow F=qE \\\ \Rightarrow F=2\times {{10}^{-9}}\times 9\times {{10}^{6}} \\\ \therefore \overrightarrow{F}=0.018\widehat{n}N/C

Hence, the force experienced is 0.018n^N/C0.018\widehat{n}N/C.

Note: A stationary charge produces electric field around itself. The extent up to which the influence of the field can be felt depends up on the magnitude of the charge. We can make use of Coulomb’s law to find out electric fields or electric force. If we know the field at any point then we can easily find out the electric force. Also, like the electric force, the electric field is too a vector quantity.